Question: The lifespans of porcupines in a particular zoo are normally distributed. The average porcupine lives $21.6$ years; the standard deviation is $4$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a porcupine living less than $29.6$ years.
Solution: $21.6$ $17.6$ $25.6$ $13.6$ $29.6$ $9.6$ $33.6$ $95\%$ $2.5\%$ $2.5\%$ We know the lifespans are normally distributed with an average lifespan of $21.6$ years. We know the standard deviation is $4$ years, so one standard deviation below the mean is $17.6$ years and one standard deviation above the mean is $25.6$ years. Two standard deviations below the mean is $13.6$ years and two standard deviations above the mean is $29.6$ years. Three standard deviations below the mean is $9.6$ years and three standard deviations above the mean is $33.6$ years. We are interested in the probability of a porcupine living less than $29.6$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $95\%$ of the porcupines will have lifespans within 2 standard deviations of the average lifespan. The remaining $5\%$ of the porcupines will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({2.5\%})$ will live less than $13.6$ years and the other half $({2.5\%})$ will live longer than $29.6$ years. The probability of a particular porcupine living less than $29.6$ years is ${95\%} + {2.5\%}$, or $97.5\%$.